Topic 5 Test Review MARK SCHEME

1. Which processes are exothermic?

I.

II.

III.

A. I and II only

B. I and III only

C. II and III only

D. I, II and III

Markscheme

D

Examiners report

Students found this question also to be difficult with 35.83% correct answers. C was by far the most common answer suggesting that students were aware of combustion being an exothermic process but not neutralization.

5a. The enthalpy changes of three reactions are given below.

What is the enthalpy change for the following reaction?

A.

B.

C.

D.

Markscheme

D

6. The enthalpy change for the reaction between zinc metal and copper(II) sulfate solution is . Which statement about this reaction is correct?

A. The reaction is endothermic and the temperature of the reaction mixture initially rises.

B. The reaction is endothermic and the temperature of the reaction mixture initially drops.

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C. The reaction is exothermic and the temperature of the reaction mixture initially rises.

D. The reaction is exothermic and the temperature of the reaction mixture initially drops.

Markscheme

C

Examiners report

Answer D was the most common error.

9. Consider the following two equations.

What is , in kJ, for the following reaction?

A.

B.

C.

D.

Markscheme

A

Examiners report

There was concern about the use of algebraic notation rather than actual numerical data. This has been used since November 2010 so candidates should be familiar with this type of question. (In fact, some G2s in the past have suggested it would be better to use algebraic notation!) In the event it was the sixth easiest question; over 81% of candidates gave the correct answer and 9% gave B.

16. Identical pieces of magnesium are added to two beakers, A and B, containing hydrochloric acid. Both acids have the same initial temperature but their volumes and concentrations differ.

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Which statement is correct?

A. The maximum temperature in A will be higher than in B.

B. The maximum temperature in A and B will be equal.

C. It is not possible to predict whether A or B will have the higher maximum temperature.

D. The temperature in A and B will increase at the same rate.

Markscheme

A

Examiners report

One G2 comment stated although this was a good question, it would be challenging for many SL candidates. In fact, although this was the fifth hardest question on the entire paper, 43% of candidates still managed to get the question correct.

27. A student measured the temperature of a reaction mixture over time using a temperature probe. By considering the graph, which of the following deductions can be made?

I. The reaction is exothermic.

II. The products are more stable than the reactants.

III. The reactant bonds are stronger than the product bonds.

A. I and II only

B. I and III only

C. II and III only

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D. I, II and III

Markscheme

A

Examiners report

There were two G2 comments on this question, both suggesting that the graph given was confusing to candidates. In this question candidates had to use a combination of ideas to ascertain that the correct answer is A, namely I. and II. From the graph shown, candidates need to realise that the reaction is exothermic, and therefore from this information, the products are more stable than the reactants. 55% of candidates got the correct answer.

29. Consider the following enthalpy of combustion data.

What is the enthalpy of formation of ethane in ?

A.

B.

C.

D.

Markscheme

C

Examiners report

There were three G2 comments on this question on Hess’s law, all of which stated that giving x, y and z variables instead of numeric data was confusing. However, candidates do not have the use of a calculator in P1 and hence it is common practice to use algebraic notation for this purpose. This notation has been used previously in P1 (though not always). In addition, this is a very common question and in fact, candidates had no problem whatsoever answering this question, with 80% getting the correct answer, C. The question was the third easiest question on the paper.

26. C

52a. [4 marks]

The standard enthalpy change of three combustion reactions are given below.

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Determine the change in enthalpy, , in , for the formation of propane in the following reaction.

Markscheme

\(\Delta H =+ {\text{2219 (kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}}{\text{)}}\);

: ;

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: ;

;

Award [4] for correct final answer.

Examiners report

In contrast, question 2 a) which involved Hess’s Law calculation, was answered correctly by candidates of all capabilities.

52b. [1 mark]

A catalyst provides an alternative pathway for a reaction, lowering the activation energy, . Define the term activation energy, .

Markscheme

minimum energy needed (by reactants/colliding particles) to react/start/initiate a reaction / for a successful collision;

Allow energy difference between reactants and transition state.

Examiners report

The definition of activation energy in part b) was reasonably well answered, with some candidates losing marks for omitting the word minimum from their response. However, it is disappointing that even very good candidates sometimes fail to score marks for definitions.

52c. [3 marks]

Sketch two Maxwell–Boltzmann energy distribution curves for a fixed amount of gas at two different temperatures, and and label both axes.

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Markscheme

x-axis label: (kinetic) energy/(K)E and y-axis label: fraction of molecules/particles / probability density;

Allow velocity/speed for x-axis.

Allow frequency / number of molecules/particles or (kinetic) energy distribution for y-axis.

correct shape of a Maxwell–Boltzmann energy distribution curve;

Do not award mark if curve is symmetric, does not start at zero or if it crosses x-axis.

two curves represented with second curve for to right of first curve, lower

peak than first curve and after the curves cross curve needs to be above curve;

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Examiners report

Several candidates sketched very clear, correct Maxwell-Boltzmann curves in part c). Most scored at least 1 mark for this question. Some did not know what labels to put on the axes. Some did not realise that the area under the curves represents the total number of particles so as temperature increases the peak of the curve shifts to the right and is lower than the peak at the lower temperature.

53h. [3 marks]

Chloroethene, HC=CHCl, the monomer used in the polymerization reaction in the manufacture of the polymer poly(chloroethene), PVC, can be synthesized in the following two-stage reaction pathway.

Determine the enthalpy change, , in , for stage 1 using average bond enthalpy data from Table 10 of the Data Booklet.

Markscheme

Bonds breaking:

1 (C=C) 4 (C–H) 1 (Cl–Cl)

Bonds forming:

1 (C–C) 4 (C–H) 2 (Cl–Cl)

Enthalpy change:

Sum of bonds forming – Sum of bonds breaking = 2,512 – 2, 650 = -138 kJ per mol;

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Award [3] for correct final answer.

Examiners report

In part c) (ii) the calculation of using bond enthalpies was done well. Some candidates failed to use the C=C bond enthalpy value and some did not recall that bond breaking is endothermic and bond formation exothermic.

53i. [1 mark]

State whether the reaction given in stage 1 is exothermic or endothermic.

Markscheme

exothermic;

Do not award mark unless based on some value for part (iii).

Examiners report

Nearly everyone scored a mark in c) (iii) as follow-through marks were awarded.

2. C(s) + 2F2(g) → CF4(g) ∆H1 = –680 kJ;4F(g) → 2F2(g) ∆H2 = 2(–158) kJ;C(g) → C(s) ∆H3 = –715 kJ;

Accept reverse equations with +∆H values.

C(g) + 4F(g) → CF4(g) ∆H = –1711 kJ,

so average bond enthalpy = 41711–

= –428 kJ mol–1 or +428 kJ mol–1 4Accept + or – sign.Lots of ways to do this! The correct answer is very different from the value in the Data Booklet, so award [4] for final answer with/without sign units not needed, but deduct [1] if incorrect units. Accept answer in range of 427 to 428 without penalty for sig figs.If final answer is not correct use following;Award [1] for evidence of cycle or enthalpy diagram or adding of equations.Award [1] for 2F2 (g) → 4F(g) 2×158 seen.

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Award [1] for dividing 1711 or other value by 4.[4]

4. (a) exothermic because temperature rises/heat is released; 1

(b) to make any heat loss as small as possible/so that all the heat will be given out very quickly; 1

Do not accept “to produce a faster reaction”.

(c) heat released = mass×specific heat capacity×temp increase/q = mc∆T =/100×4.18×3.5;= 1463 J/1.463 kJ; (allow 1.47 kJ if specific heat = 4.2)amount of KOH/HCl used = 0.500×0.050 = 0.025 mol;∆H = (1.463÷0.025) = –58.5 (kJ mol–1); (minus sign needed for mark) 4

Use ECF for values of q and amount used.Award [4] for correct final answer.Final answer of 58.5 or +58.5 scores [3].Accept 2,3 or 4 significant figures.

(d) heat loss (to the surroundings);insulate the reaction vessel/use a lid/draw a temperature versus time graph; 2

(e) 3.5°C/temperature change would be the same;amount of base reacted would be the same/excess acid would not react/KOH is the limiting reagent; 2

[10]

6. (i) energy required to break (a mole of) bonds in the gaseous state/energy given out when (a mole of) bonds are made in thegaseous state;average value from a number of similar compounds; 2

(ii) (HӨreaction = (∑BEbreak BEmake))

= [(837) + 2(436)] [(348 + 4(412)];

= 287(kJ/kJ mol1); 2Award [1 max] for 287 or + 287.

(iii) (BE): CCl > CBr > CI/CX bond becomes weaker;halogen size/radius increases/bonding electrons further away fromthe nucleus/bonds become longer; 2

[6]

7. (a) (i) standard enthalpy (change) of reaction;(temperature) increase;reaction is exothermic/sign of H is negative; 3

(ii) more (negative);heat given out when gas changes to solid/solid has less enthalpy thangas/OWTTE; 2

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(iii) –389 kJ; 1[6]

8. (i) it is an element/no other species with just a Br-Br bond/OWTTE; 1

(ii) (sum bonds broken =) 412 + 193 = 605;(sum bonds formed =) 276 + 366 = 642;(H =) –37 kJ; 3

Award [3] for correct final answer.Award [2] for “+ 37”.Accept answer based on breaking and making extra C-H bonds.

(iii)

Enthalpy CH4 + Br2

CH3Br + HBr ;

2Award [1] for enthalpy label and two horizontal lines, [1] for reactants higher than products.ECF from sign in (iii), ignore any higher energy level involving atoms.

(iv) (about) the same/similar;same (number and type of) bonds being broken and formed; 2

[8]

12. (a) (Amount of energy required to break bonds of reactants)

8×412 + 2×348 + 612 + 6×496/7580 (kJ mol1);

(Amount of energy released during bond formation)4×2×743 + 4×2×463/9648 (kJ mol1);

H = 2068 (kJ or kJ mol1); 3ECF from above answers.Correct answer scores [3].Award [2] for (+)2068.If any other units apply 1(U), but only once per paper.

(b) exothermic and HӨ is negative/energy is released; 1Apply ECF to sign of answer in part (a).Do not mark if no answer to (a).

[4]

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54a. [1 mark]

If white anhydrous copper(II) sulfate powder is left in the atmosphere it slowly absorbs water vapour giving the blue pentahydrated solid.

It is difficult to measure the enthalpy change for this reaction directly. However, it is possible to measure the heat changes directly when both anhydrous and pentahydrated copper(II) sulfate are separately dissolved in water, and then use an energy cycle to determine the required enthalpy change value, , indirectly.

To determine a student placed 50.0 g of water in a cup made of expanded polystyrene and used a data logger to measure the temperature. After two minutes she dissolved 3.99 g of anhydrous copper(II) sulfate in the water and continued to record the temperature while continuously stirring. She obtained the following results.

Calculate the amount, in mol, of anhydrous copper(II) sulfate dissolved in the 50.0 g of water.

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Markscheme

;

Examiners report

Question 1 was a generally difficult question for candidates, but most students did pick up marks thanks to the application of error carried forward (ecf). In part (a) students could usually calculate the moles of anhydrous copper sulphate.

54b. [2 marks]

Determine what the temperature rise would have been, in °C, if no heat had been lost to the surroundings.

Markscheme

26.1 (°C);

Accept answers between 26.0 and 26.2 ( °C).

temperature rise (°C);

Accept answers between 6.9 °C and (7.1 °C) .

Award [2] for the correct final answer.

No ECF if both initial and final temperatures incorrect.

Examiners report

Very few candidates could correctly extrapolate the graph to calculate a temperature rise of 7.0 ºC.

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54c. [2 marks]

Calculate the heat change, in kJ, when 3.99 g of anhydrous copper(II) sulfate is dissolved in the water.

Markscheme

heat change ;

Accept 53.99 instead of 50.0 for mass.

;

Allow 1.6 (kJ) if mass of 53.99 is used.

Ignore sign.

Examiners report

Calculating using also caused problems as many students used the mass of the copper sulphate instead of the mass of water, and some also added 273 to the temperature change. Many candidates also forgot to convert to kJ.

54d. [1 mark]

Determine the value of .

Markscheme

;

Value must be negative to award mark.

Accept answers in range –58.0 to –60.0.

Allow –63 (kJ mol) if 53.99 g is used in (iii).

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Examiners report

The last part of this question required the calculation of , here many students forgot the – symbol to indicate it was exothermic and so did not gain the mark.

54e. [2 marks]

To determine , 6.24 g of pentahydrated copper(II) sulfate was dissolved in 47.75 g of water. It was observed that the temperature of the solution decreased by 1.10 °C.

Calculate the amount, in mol, of water in 6.24 g of pentahydrated copper(II) sulfate.

Markscheme

;

.

Examiners report

In part (b) the problems were similar as students used incorrect values in their calculation but were able to obtain some marks by error carried forward.

54f. [2 marks]

Determine the value of in .

Markscheme

;

;

Accept mass of 47.75 or 53.99 instead of 50.00 giving answers of +.8.78 or +9.9.

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Do not penalize missing + sign but penalize – sign unless charge already penalized in (a) (iv).

Examiners report

In part (b) the problems were similar as students used incorrect values in their calculation but were able to obtain some marks by error carried forward.

54g. [1 mark]

Using the values obtained for in (a) (iv) and in (b) (ii), determine the value for in .

Markscheme

Examiners report

In part (b) the problems were similar as students used incorrect values in their calculation but were able to obtain some marks by error carried forward.

54h. [1 mark]

The magnitude (the value without the or sign) found in a data book for is .

Calculate the percentage error obtained in this experiment. (If you did not obtain an answer for the experimental value of then use the value , but this is not the true value.)

Markscheme

;

If 70.0 kJ molis used accept 10.3%.

Examiners report

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In part (c) many could calculate the % error and apply Hess’s law to calculate . Throughout this question there were numerous instances of students using an incorrect number of significant figures and this led to another mark being lost.

54i. [2 marks]

The student recorded in her qualitative data that the anhydrous copper(II) sulfate she used was pale blue rather than completely white. Suggest a reason why it might have had this pale blue colour and deduce how this would have affected the value she obtained for .

Markscheme

the anhydrous copper(II) sulfate had already absorbed some water from the air / OWTTE;

the value would be less exothermic/less negative than expected as the temperature increase would be lower / less heat will be evolved when the anhydrous salt is dissolved in water / OWTTE;

Do not accept less without a reason.

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